k>lF%fk ■ ^B 



ANSWERS TO PROBLE: 

FOR THE 

SECOND EDITION 

OF 

DADOURIAN'S 

Analytical Mechanics 



BT 



H. M. DADOURIAN 

Associate Professor of Physics, Trinity College, 
Hartford, Conn. 




NEW YORK 
D. VAN NOSTRAND COMPANY 

Eight Warren Street 
H^l H 1921 



ANSWERS TO PROBLEMS 

FOR THE 

SECOND EDITION 

OF 

DADOURIAN'S 

Analytical Mechanics 



BY 



H. M. DADOURIAN 

Associate Professor of Physics, Trinity College, 
Hartford, Conn. 



NEW YORK 
D. VAN NOSTRAND COMPANY 

Eight Warren Street 
1921 






{ 



> 



Copyright, 1921, 

BY 

D. VAN NOSTRAND COMPANY 



OCT- 25 1921 



TECHNICAL COMPOSITION CO. 
CAMBRIDGE, MASS., U. S. A. 



©GU627413 



ANSWERS TO PROBLEMS FOR THE SECOND 
EDITION OF DADOURIAN'S ANALYTICAL 
MECHANICS 

Instructors may be divided into two groups according to 
their attitude toward " answers." The group which is in 
favor of answers claims that these encourage the student 
to work out problems. The group which is opposed to 
them claims that they serve to shift the emphasis from the 
proper solution of the problem to " getting the answer" 
and discourage the development of the student's confidence 
by making the agreement of his result with the answer the 
criterion for the value of his work. 

I have in the past resisted the temptation of providing 
my book with answers in spite of many calls for them with 
promises of a wider field for the book. In reconsidering 
this question, however, I came to the conclusion that the 
objections to answers do not hold good in the case of the 
second edition of my " Analytical Mechanics." In this 
edition, directions are given for a general method for solv- 
ing problems and this method is illustrated by many ex- 
amples scattered through the book*. If the instructor would 
insist upon the student's following this method, he would 
find that not only does the student's interest shift from 
"getting the answer" to solving the problem, but he becomes 
orderly, logical and efficient in his procedure. The term 
"insist" is used advisedly. The instructor will find that 
the student resists, at least during the first week or two of 
the course, any attempt to make him follow any orderly 

* Cf . Pamphlet by the author on " A General Method for Working Out 
Problems in Mechanics," D. Van Nostrand Co., New York, 1919. 

3 



4 ANSWERS TO PROBLEMS 

procedure in working out problems. This is not surprising, 
because his previous training has accustomed him to work 
carelessly, without plan or order. 

The answers to the general problems at the end of the 
chapters and to problems on trusses in Chapter IV are not 
given in the following list. These are omitted in order to 
provide the ambitious student with a means for developing 
his self-confidence and testing his ability. 

The list has been prepared in an attempt to have it ready 
within a limited time. Consequently it is to be expected 
that many errors have crept in. I hope to eliminate these 
before incorporating the list in the third edition of my 
Mechanics, which is shortly to be brought out. I should 
be greatly obliged if those who use the book would call my 
attention to errors. 

H. M. Dadourian 



ANSWERS TO PROBLEMS 

Page 8 
1. R = 39.8 cm., = - 27°.6. 2. R = 17.8 cm., = - 85°.6. 

Page 13 

1. b = a V3. 2. c = 2a, = 60°. 3. R = 0. 6. (a) 7.5 cm.; (6) 13 
cm.; (c) - 15 cm.; (d) - 15 cm.; (e) 0.0; (/) 7.5 cm. 7. a = 12.5, = 
53M3. 8. (1) R = 21.0, = 87°.3; (2) P = 8.06, = - 6°.96; (3) P = 
13.04, = - 41°.2; (4) R = 12.65, d = 108°.4. 

Pages 24-27 

W I 
3. F = 259.8 lbs. 4. T 7 =y - , • 5. Ti = 40 lbs., T 2 = 30 lbs. 

10. SW t = 17W*. 11. F = 3.9 tons, T = 3.6 tons. 12. W = , \ 2 a • 

13. (a) F = 625 lbs.; T = 325 lbs.; (6) F = 693 lbs.; T = 400 lbs. (c) F = 
1000 lbs., T 7 = 1414 lbs. 14. (a) F= 2500 lbs., T= 500 lbs.; (6) F = 1639 lbs., 

T 7 = 346 lbs.; (c) F = 1800 lbs., T 7 = 60C lbs. 16. F = f • 17. (a) 0.87; 

2 tan 

(6) 2.83; (c) 5.72; (d) 28.65; (e) 57.3; (/) 2S6.5; (a) 2865; (h) oo. 

Pages 31-32 

1 d iQooiu co o . \ sin a sin a , n l + sin a 
1. P = 132.2 lbs., // = 5.2. 2. ^a) r — : , or, (o) , 

1 + COS a 1 — COS a COS a 

1 — sin a _ u cos a — sin a 

or, — — • 3. ix = 1. 4. P = TF , =r , = tan-V- 6. P = 

' cos a . V 1 + M 2 

30 lbs., R = 80.8 lbs. . 7. P = 27.85 lbs., - 21°.8. 8. P = 

W vV + 2(l -cosai/xsina), tan = ] ~ COSQ: . 9. = 18°.3. 10. \ tana, 

sm a ± fj, 3 

5 
or, - tan a, or, 3 tan a. 11. 91 lbs, 12. 58.5 lbs. 13. 123.5 lbs. 

14. 175 lbs. 15. 375 lbs. 

Pages 33-34 
1. R = 71.7 lb., = 7°.S. 4. P = P VI, ai = « 2 = a - 54°.7. 



6 ANSWERS TO PROBLEMS 

Page 45 
1. G = 0, or G = 2 Gi. 2. G = 0, or G = 60 lbs. ft. 3. G = 3 Gi. 

4. G = 120 lbs. ft. 5. G = Wr ^— • 

2 sin a 

Pases 53-56 



2. 20 lbs., 15 lbs. 3. Rx = TT Vl + M 2 , = tan~V, #2 = nW, d 2 = 0. 

4. 1200 lbs. 5. 20.3 ft. 6. 5.8 in. 7. T = P = W 9 * =• 9. (a) 

1 v p — a 2 

T = 1042 lbs., R = 914 lbs., = 65°.8. (b) T = 2708 lbs, R = 1256 lbs., 
= 22°.3; (c) R = 2.13 tons, T = 3.54tons, 6=4°.3. 10. -■ 11. (a) 353.7 lbs.; 
(b) 250 lbs.; (c) 191.41bs.; (d) 151.21bs. 12. F = 1.25 tons, 72 = 1.953 tons, 

= 50°.2. 13. — — ■ 14. h = 6.43 ft., k = 8.57 ft. ' 15. 0.577. 

• a cos a 

PI 

16. F = I r; ^ — / . - 1 where I = 30 in., a = 1.5 in, Z> = 

tan (0 - 0) Va 2 + b 2 - 2a6 cos 

1 in., <£ = angle I makes with vertical, 6 = angle the radius to point of contact 
makes with I; a cos d — a sin 6 cot 4> = b. 17. Let <f> = 90° .'. a cos d = h, 

tan (i/f - 6) = cot 6 , : . F = 920 lbs. 

V a 2 — o 2 

Pages 60-61 

.. 2 1 

1. R = SP, G = 3 Pa, a = side of square. 2. R = 3 P, G = — ?= Pa. 

3. # = 0, G = Pa. 6. Pi + P 2 + P 3 = O. 

Page 63 
11 3 3 

1. TTflP, TT/JLOP. 2. ~ TTuP, jZ -KllOP. 3. '- TTfjlP, 5 TfJiaP. 

Page 65 

L !„£^p. 2. IAp. 3. MLr 4. |4*-p. 

3 6 2 — a 2 3 sin a 3 sin a: 3 sine* 



6. ,ap(^- - 5*2). 

\sin 2 « sin a/ 



Pages 76-77 



In the following answers the external forces are given first, The hori- 
zontal and vertical components of the reaction at the base are given 
(a) 643, 643, 1500, 2250, 375, and 1949 lbs. 
(6) 2.77, 1.3S, 4.40, 0.4, 2.4, and 3.4 tons. 



ANSWERS TO PROBLEMS 7 

(c) 1050, 1050, 1750, 1617, and 2914 lbs. 

(d) 974, 974, 2000, 1000, 433, and 500 lbs. 

(e) 1.886, 1.866, 2.00, 0.80, 3.80, and 3.96 tons. 
(/) 1.5, 1.21, 0.25, and 1.32 tons. 

(</) 1250, 250, 1250, 1250, 1274, and 1768 lbs. 
(h) 1.5, 14.5, 22.5, 1.54, 3.59, 1.37, and 17.2 tons. 
(?0 12.25, 18.86, 18.86, 12.8, 1.73, and 5.87 tons. 

Page 109 

600 tons and 671 tons. 

Page 113 

3. 1.65 ft. and 1 mile. 4. 1 mile and 40 miles. 6. 154 and 184 tons. 
7. 506.5 ft. 

Pages 118-119 

1. 41 lbs. 2. 4630 lbs. 3. (a) .0039 W; (6) 2576 W. 4. 8280 lbs. 

6. (a) 175 lbs. (d) 528 lbs. ' (g) 281 lbs. (j) 2536 lbs. 

(5) 223 lbs. (e) 491 lbs. (h) 2536 lbs. (k) 8874 lbs. 

(c) 331 lbs. (/) 385 lbs. (*) 7640 lbs. (Z) 36860 lbs. 

Pages 118-119 

6. (a) 221 and 771 lbs. (e) 649 and 1149 lbs. 
(6) 228 and 778 lbs. (f) 318 and 818 lbs. 

(c) 420 and 920 lbs. (g) 131 and 631 lbs. 

(d) 152 and 652 lbs. (h) 175 and 675 lbs. 

7. Answers follow immediately from those of preceding example. 

Page 129 
1. (a) y = - x, v = Va 2 + 6 2 ' 

(6) *y = x - ^x\ v = V«2 + (a - gt)\ 

(c) y = b cos - x, v = va? + b 2 o* 2 sin 2 cof. 

(d) x = a sin - y, v = Vfr 2 -|- a 2 w 2 cos 2 co£. 

(e) a 2 x 2 + b 2 y 2 = a 2 6 2 , v = aco. 
CO y = - x, v = w Va 2 + & 2 cos 
(gr) #?/ = a 2 , v = &Va; 2 -J- ?/ 2 . 



8 ANSWERS TO PROBLEMS 

Page 133 

1. 50*^ , 6 = 36°.9. 2. 528 ft, 3. (a) =-^-r v, (b) -J^—r v. 4. v u = 
nr. tl — h ti — h 

2 v sin °- . 5. 38.8 "^- S , N 21°.7 E. 6. 21 ^^ , 7/. 45° E. 7. v 12 = 2t; sin 
2 hr. hr. 

^±-\ 8.2672°-^. 9. .0224^15. 
4 sec. hr. 

Pages 136-137 

1. .0169 and .0175 radians per sec. 2. 29°.6. 3. 17.81 ^^ , 400 w I ^— 

hr. min. 

. _ _ . rads. „-. cms. nM cms. _ v „„ „ _ _.. , miles _ y 

4. 8.04 , 301.3 , 301.3 . 5. - cos 2 0. 6. 0.29 cos X . 8. - , 

sec. sec. sec. a sec. a 

v _ b sin <b a + b cos © „•_ . a -\- b cos d> 

-,9. wr = r^f, »p = — ! — : -v. 10. = — ! -v. 

a r 2 r 2 r 

Page 139 

1. 30.5 — : . 2. 32.15 — , • 7.89 X 10 4 ^f . 3. 6.94 — 1 . , 0.228 — , , 
sec. 2 sec. 2 hr. 2 sec. 2 sec. 2 

900 EH • 



.. ^y 



■ P. ^^ 2 

"•" 2z 2x V4.T 2 + 2 



Page 144 



px 



-2. (a) Path, w 2 = b 2 x: x = a; 2/ = 1 &£-£ ; x = 0; i/ = — 1 

a 2 4^ 

2. (6) Path, gr.-r 2 — 2 a6 x = 2 a 2 ?/; x = a; ?/ = 6 — <^; $ = 0; y = — 0. 
v = vV + (6 ~^p. f = -g. 

2. (c). Path, a?/ = bx; x = /cae**; ?/ = Arbe 1 *; a! = A- 2 ae^; ?/ = A:*6e*'; 

2. (d). Path, ?/ = be H; x = a; y = — kber-**; x = 0; ?/ = h 2 ber-*t] 
v = Va 2 + AWe-**; / = #%«-*. 

2. (e). Path, 7/ = 6 sin — ; x = a; y — bcocoscot; x = 0; y = — 6co 2 sinco^ 

y = Va 2 -f- 6 2 co 2 cosW; f = — bw- sin co£. 

2. (/). Path, x = a cos -^ ; x = — aco sin o>£; ?/ = b; x = ao 2 cos cot; 



/ v 2 

'/'/ =0; v = v 6 2 -f- a 2 a> 2 sin 2 «£; / = aw 2 cos wJ. 3. ar, a 2 r. 4. — cos cot, 



y 2 y 2 y 2 y 2 

- sin cot, 0,0. — , — , — 
a a a a 



ANSWERS TO PROBLEMS 9 

Page 146 

1. 0.218 =: . 2. 10.9. 3. 5.24 sees. 4. (a) cob cos oot, - oo 2 b sin at: 

sec 2 . ' 

(6) — cob sin o>£, — co 2 6 cos cot; (c) b, 0; (d) b, 0. 5. (a) p = =- 0; (6) p 2 = 
a 2 f 1 — j- J ; (c) p = a sin ^ d; (d) p = ae*. 

Pages 165-166 

2. 6.21 lbs., 250 ft. 3. 4.5 sec. 4. 5.01 sec. 5. 1.525 X 10 5 — • 6. 

sec. 2 

50.96 m., 10.2 sec, 7. 9.31 lbs. 8. 399.7 lbs. 9. 2497 lbs. 10. 5.4 sec. 

11. 62100 ft, 12. 484 ft. 13. 176 — . 14. 23.1— 15. 1 sec. 

sec. sec. 

Page 167 
2. h = -7= • 3. vo 2 - v ' 2 = 2gs sin a. 5. i/aZ^Z-A 2 Y 

Page 168 

1. 6.37 miles. 2. 112.7— . 3. (a) 161.1 — ; (6) 64.5 — - 

sec. sec. ' sec. 

Page 174 

ft. 
1. 185 — '-, 48°.8. 2. 53.3 miles (effect of elevated position is negligible). 

SCO* 

3. i A 4. 3.827 X 10 8 dynes form = 5gm. 5. (a) 52.1 ^£? ; (6) 67.3 ^^- 

6. 94.7 ft. 7. 36.2 ft. 10. 100.7 m. 

Pages 176-177 

mvv 2 miles A K 

1. — = m 2 ^. 2. 4.81 — — • 3. 5.6 tons. 4. P = 2tt \J- • 6. TaZ. 

t sec. j q 

7. AW = 47r2 T !f ig cos 2 a. 8. 0.009 — - • 10. 1.41 hr. 11. 0.27 per cent. 

F 2 sec 2 

13. 8.2 in. 

Page 179 
1. 11.63 min. 2. t a d 2 . 

Pages 183-184 



1 / kVn\ 1 

1. 6.8 m\ 2. 6.0 w*. 5. B - ^ log ^1 + - ), T = ^= X 



10 ANSWERS TO PROBLEMS 

Page 186 



2. 4.92 mileS 



sec. 



Page 196 



4 _ 2 _ 7r ira 9 . , , 9 . , . . 4a 4a . 4a 46 
1. 0, 3a. 2. -, T . 3. ggalM, ^a.M. 4. -, - . 5. -, — 

, 5a A „ 2 sinf) 2 l-cos# . - - / 4(6 3 - a 3 \ 

Page 197 

* 3 , n „ 5 n «> 3a >i 3 a k i """ « 3 6 2 — a 3 _ 

1. j*, 0. 2. -a,0. 3. -g-,0. 4. -a,0. 5. 1, - • 6. g 55-^,0. 

199 

„ a sin a(l — cos 20) . 4a 

L -2F' 20 *•"• I" 

Page 201 
1. |ft,0. 2. §«,0. 3. £+|i,0. 4. § a, § 6. 6. * a, 0. 

Page 203 

7 1Q 1Q /r n Z r 

1. (a) 0, i a; (6) ^7 a, ^ a; (c) 0, 0.64a: (a 1 ) £ , 0; (e) 0, 0.57a. 2. - 



4 -, v-/ 2 l w, 21 ,N ' ' ' v '3» w 'w w »"«"- - 53_ a s> 

o n ^ >■ n 3 a 2 - ft 2 . n fei + ft* ft 2 

3 °« 8* 4 - °' ~4a^T * 6 " °' -4^' 6 - 5 °' °' 

Page 205 

« 3 V§ „ 3 , 2a 2« 3a 3a 

1. 4 7r 2 ac. 2. 4 7r 2 a 2 c. 3. -f2 vra 2 . 4. tt; tto 3 . 6. — , — . 6. j- , -t- ■ 

4 lO 7T 7T 47T 4tT 

Page 207 

. ma 2 m& 2 m a 2 ?) 2 . wa 2 _ rap 8 , .31 ,, 11 

1- X- 2 - T" 3 ' 6" tf + * 4 - T" 5 - 2 * 6 (a) 30 m6 °' 15 ™ ; 

/n 53 M 1 o ^ 53 , 2 29 , , „ 55 M 79 . 59 ,. 37 _ 
(6) ^ mb\ ^ ma 2 ; (c) ^ ra6 2 , ^ ma 2 ; (d) — rab 2 , — ma 2 , ^ ra6 2 , — raa 2 . 

Page 215 

1. (a) | (a 2 + 6 2 ); (6) ^ (3 a 2 + ¥); (c) 2 (a 2 + 6 2 + 0. 

2. (a) 5 (a 2 + 6 2 ), 2 (a 2 + ?) • 3. = (a 2 + 6 2 ), (6) | («« + J) . 



ANSWERS TO PROBLEMS 11 

. ' v m/abfa 2 4-fc 2 )-a'b'(a' 2 4 6' 2 \ ,. , m /a% - a' 3 b' , l 2 \ c , , ma 2 
4 " (a) 3 ( a6-a'6' ' ) ; {b) 3 (l*=SP + 4> 6 (n) ~12 ' 

(6) ma 2 4- \ m 4— T5 " 8 - i m « 2 - 9 - ( a ) 77^ ™« 2 > 4 m (« 2 + 4 /i2 )- 10 - k ma > 2 > 
oi, I ma 2 . 11. (a) \ mpa, (b) I ma (2 p + 3a). 12. (a) ™ (a 2 4- 6 2 ), 

O a O O 

(6) ^ (6 a 2 4- ¥). 
o 

Page 229 
.1. 0.737 ft. lb. 2. 1.356 X 10 7 ergs. 3. w(Jk 4- n VjT^h 2 ). 4. 1205 
kg.m., 205 kg.m. 5. 1563 kg.m. 6. 2mgh, irmgh. 7. 9250 ft. lbs. 8. 1.11 
X 10 5 lbs. 9. (a) 255 sec, 3.36 X 10 7 ft. lbs.; (6) 580 sec, 7.65 X 10 7 ft. lbs.; 
(c) 178 sec, 3.36 X 10 7 ft. lbs.; (d) 281 sec, 5.41 X 10 7 ft. lbs.; (e) 478 sec, 
2.36 X 10 7 ft. lbs.; (/) 3850 sec, 5.08 X 10 8 ft. lbs. 10. 1.056 X 10 8 ft. lbs. 
11. 1.431 X 10 8 joules. 



1. 66.3 lbs. 3. 1.125 ft. lbs. 4. \kd\ 5. 4.8 ft. lbs. 6. 15 lb., 45 lbs. 



Pages 231-232 
1 

2' 

Pages 339-340 

1. ML- x T-\ 0, ML~ l T~*. 2. 3.18 X 10 7 fe 3. 9.54 lbs. 4. 1. 

in. 2 

6. 9.56 lbs. ft. 6. 356 lbs. 7. The semi- vertical angle is given by L sin a — 

2d cos a 4- L sin 2a = 0. 9. 6000 dynes. 10. 0.556. 12. ^ = 1 - -.. — = 

K a 4 m 

1 --. 13. 1. 14. 1. 



Pages 243-244 

L F - »4i- 2 - (a) I i ™ I ■• (c) S ; (d) 2 - » 4 = <" 7 = W 6. 

3 . w . 4 . sZ= _4L_, r = _f_ ; 6 . (a) *L; <»»=». 

2 7r tan a 2 7r tan a it — r rp 



Pages 250-251 

; - 

S€ 

7. 3.6 X 10 6 lbs. 9. 788 ft. 



2. 22250 dynes. 3. 783 — • 4. 5.26. 5. 1698 lbs. 6. 53.6 — 

sec. sec. 



12 ANSWERS TO PROBLEMS 

Page 254 

1. 13.7 *5I: 2. 1882 ft. lbs. 4. 26.9 tons. 6. 6.16 X 10 36 calories. 

see. 

TPV 

6. 2.01 X 10 11 mile 3 . 7. 3.27 ft. lbs. 8. 39.1 — . 9. 2.21 X 10- 3 lb. ft., 

sec. 

2349 revs. 10. (a) KFl; (b) i KFl 2 . 

Pages 256-257 

2. 33.3 I1 ^ S . 3. 14.4 H.P. 4. 51.2 H.P. 5. 124 K.W. 6. 208 H.P. 

hr. 

lbs 

7. 1222 lbs. 8. 97.3 r^f. 9. 4.78 X 10 8 K.W. 10. 23.5 H.P. 11.18.8 



ID 



K.W. 12. 1762 lbs. ft., 2.367 X 10 7 dynes cm. 13. 2.65 X 10 9 dynes cm. 

14. 11.9— — • 15. 119.2 ton vds. 16. 128.8 ton yd?., 105 tons, 
sec. 

17. 0.22 H.P. 18. 1.74 H.P. 19. 150 H.P. 20. 166°.2. 21. 29920 tons. 

22. 1414—- 23. 1.5 tons. 24. 240 H.P 25. 10 lbs. per ton. 26. 2070 lbs. 
sec. 



Pages 265-266 



miles . k y§^ _ 27r 2 n 2 ma- 



27. 980 lbs. 



1. 3.75 X 10 7 ft. lbs. 2. 22.5 ft. 3. 7.1 ~^~ • 4. y ^ . 5. 

6. Vga, ^5ga. 8. cos « = - . 9. 3 cos 6 = 1 + 2 cos a. 10. Vgl • 

o 

11. I UK 12. I k6?. 13. 36.3 — 14. (a) 44.5 — ; (b) 35.7 — ; (c) 39.8 — • 
2 2 sec. sec. sec. l sec. 

15. 11.1ft. 16. (a) 5.55 ft.; (6) 5.48 ft.; (c) 7.40 ft. 17. 802—- 

sec. 

Pages 282-283 
« n . m m : 3 m(b 2 - a 2 ) , Q . ro(36V - r» -2a 8 ) 

l. (i) - 7— , - 7-3 ; (2; - 7 tttm — ^tj °> w ~ y 



r ' V ' v ' ' 2 (6 3 - a 3 ) ' ' v ' ' 2(6 3 - a s )r ' 

w(r 3 — a 3 ) 7T7m ,< , , irytn / .r \ 

y -JT-, tt 1 - 2. - 2~V ^a 2 + x 2 - a:), - 2 -V ( -7 1 ) 

1 (b* - a 3 )r a 2 a- \Va 2 - x 2 / 

3. - T log(l+^, -j^u- 

ym r / "V+z 

4. - a2/ 1 x va z + x 2 - x 2 + a 2 log (3 + v^M- .r 2 ) 



ANSWERS TO PROBLEMS 13 

2ymf , , ~\x+i 

—tf \J + Va 2 + x 2 - Va 2 + (x + r)*j x • 

6. -— i~ Va 2 + x 2 -x , ~ -4- 1 - , ^ • 

a L Jx a 2 L Va 2 + £ 2 Jz 



_ mi — m 2 mi — m 2 

"• : — » _ t — i^— 



Pages 297-298 

3. 17190 pd. ft. 2 , -5.58—^, 7.5 min. 4. .417—,, 9102 gm. cm 2 . 

sec. 2 sec. 2 

6. 3.23 X 10 7 pd. ft. 2 6. 1200. 7. 500 dynes. 

Page 308 

4. 7.5 sec. 11. The solid sphere will roll down an inclined plane faster. 
12. The heavier man. 

Pages 315-316 

1. (a) 0.0173 1b. sec, ^; (6). 0.0245 lb. sec, tt; (c) 0.0173 lb. sec, j- ; 

(d) 0. 2. 2 mv sin - , -(«• + 0). 4. mvz = const., -y (mt'y) = — m^. 

5. 20.4 lbs. 6. A = 90°, F = 39.9 lbs. 7. 3.667 X 10 5 lbs. 8. No. 

v + 1 
9. (n + l)y, or-^^y. 10. (a) No; (6) Yes. 12. 7.032 X 10 5 lbs. 



Pages 322-323 

• 5. 480 — 
3 sec. 



3. mi = 3 m 2 . 4. e = | • 6. 480 — • 7. d 1 = erf. 8. V = p»' = 

■i can 



V/^ 1 * = V/- ■*• *■ 22 ' 4 ^T> L6 ^T- 11- 1-12^, 63°.5. 12. (a) 2X104 
» m 2 ~ mi bec - beL - 

gm - Cm ' ; (6) 4 X 10 5 ergs. 13. 750 — , 2.2 lb. sec. 15. ^-—^ V^h, 
sec sec. 2 ' 

3+6 / 7T 

— - — v gr/i. 16. - . 17. 411 sec (hint, time between collisions = circum- 
2 o 

ference-^- relative motion.) 

Page 327 

1 1 

3. 2h sin a. 7. (a) - mu 2 (l - e 2 ) cos 2 «; (6) ^ my 2 [cos 2 a \ (1 - e 2 ) - (1 + e 2 ) M ( 

,_ , . . rt , n 1 — 2 e cos 2 « 1 + 2 e cos 2 a _ 1 30- — e) 

+ M (l + e) sm 2a . 8. — .», ' — p. 9. 5 my 2 , 

1 .r x • / 1+2 cos 2 a 1+2 cos 2 a 2 5 

(a) 0.18 ft. lbs.; (b) 0.70 ft. lbs. 10. 42.4 — , 21.2 — ■ 
v ' ' sec sec. 



14 ANSWERS TO PROBLEMS 

Pages 332-333 

1. 4.52 X 10-^- 2. 434^;. 3. 5 X 10? ^f = 725^. 4. (a) 

ft. 2 in.- cm. 2 m. 2 

305^, 135^;- 5. 1275 lbs. 6. 1275 lbs. 7. ™*M ( j _ cos d V 8 420 
in. 2 in. 2 g \ 2/ 

lbs. 9. 840 lbs. 10. 1434 lbs. 11. 1680 lbs. 12. 14.6 lbs. 13. 167 lbs. 

Pages 341-342 

2. Vertical line. 3. A P = — • P = - v^ P, (if density of dust equals 

<. ,i \ a 4 71-raa 2 _ * «.,„ w i/w P^- ft. , .„ „ 

average density of earth). 4. 5--T— , • 5. 1.316 X 10 5 , 147 lbs. 

■' Snaay 2 sec. 

6. 7.4 — • 7. 6.4 — ■ 8. (a) | Wo ; (&) 2 f 1 - f -__^__ ) mJ 

sec. sec. .5 \ 5 m + m x + ra 2 / 

, x /, 1 ma 2 . \ ./jx/i 1 ma? ■ \ 

K} V 5 ma 2 + m'(a 2 + 6 2 ) / ^ W \ 5 ma 2 + m x {a? + 6 2 ) + 2 m 2 a 2 ,/ c ° ' 

Pages 347-348 

6. (a) |^ ; (6) 0, or, ^ • 6. 4 «. 7. 5 «. 8. ?co. 
2a 5 a 5 

Pages 350-351 

1. R = mg V4O-84cOs0 + 45cos 2 0, = tan- 1 3 s * n g t 2. = 90°, 
y 6 — 7 cos0 



R« = 2m# VTo. 5. 7? = ^ Vl + 63 cos 2 0, </> = tan- 1 \ tan 0. 7. R = 



8 



m#. 



8. R = mg Vl7 - .48 cos + 35 cos 2 0, = tan- 1 _ S11 * g . • 9. R = ^ 
y 6 cos — 4 2 



sin 



V 17 — 32 cos + 15 cos 2 0, </> = — tan- 1 -r-r -r ■ (a) = 7T, i2 = 4 mg, 

4(1 — cos 0) 

= 0; (b) = 0, R = 0. 10. 22 = ^ V s in 2 + 4(27 - 5 cos 0) 2 ;</»=- 

tan- 1 2(27 ^Ycos0) ' (a) * = llm{7 ' * = ° ; (5) R = 13 ' 5 ^' * = ~ 
10°.5; (c) 22 = 16 mg, = 0. 

Pages 252-253 

i / ^ y 5 ,l^n 12 y 4 n M 12» 4 . /rxn 

1. (a) — , 7 ww; (0) 0, my, or, — — , - mv. 2. (a) ^— , - mv; (0) 0, mv, or, 
a 4 o-io o a o 

36 v 7 n v sin my 2 , . . ON . 2 y sin 2my 3 \ 

5. 6=1*. 6. na 7. 6=^a. 8. 6 = 2 a. 9. 6 = — 
3 2 ma. 



ANSWERS TO PROBLEMS 



15 



Page 367 
1. 6.02 X l(Fgm.,5.49gm./cm 3 . 2. 8.08 X 10 19 tons. 3. 3.475 X 10 5 . 



Pages 374-375 



cm. cm. 

1. 37.7— ,15.08 — 
' sec. 



sec. 
6. 0.452 sec. 



2. 2 



7T y - , o. 



3. 1 hr. 25 min. 4. 4.92 



miles 
sec. 



(a) 
8. *«. 

IT 



9- - , 

.~ 2a 



I a - 

i. 2 , 



II. 



2' 



(6) 

iff 
P ' 

7r 2 o 
~P~' 

2 cr 



III. = sin 



0, 
0, 

o, 



(c) 
8 7ra 

p2 » 

2tT 2 Q 
p2 > 

8*-o 





> 

7T 


7T 


P 2 


11. 


» 


7r ? a 

P ' 


2 7r 2 a 
P 2 


12. 


(a) 


(6) 


(c) 



o, 

o, 

2 7ra 

p2 » 



(4) 

8 7ram 
pi » 

2 7r 2 am 

p2 > 

8 7raw 
2 7r 2 am 

p2 I 
(d) 

o, 
o, 

2 7rma 

p2 » 



(e) 

4 am 
~P~' 

Am 
~P~' 

4 am 
~P~ 

7r ? am 
"P 2 " 

w 

o, 
o, 
o, 



if) 

ir a 2 m 

p2 » 

4 7r 2 a 2 m 
3P 2 ' 

ira 2 m 

"P 2 " 

4 7r 2 a 2 m 
3P 2 ; 

(/) 
7rcoa 2 m 

8 ' 
7rcoa 2 m 

2 7r 2 ma 2 
P 2 



7r a 2 ??i 

~p2-' 

2 7r 2 a 2 m 
3P 2 

7ra 2 m 
~pT 

2_7r 2 a 2 m 
3P 2 

(?) 








Pages 377-378 

The resultant motion is represented by x = a sin (wi + 5) 
The constants a and b are given in the following table, where 





I 




2 7T. i • - 

— to, o = sm 8, 
V 


c = 


cos 5 




a 2 






tan 6 


(1) 


(ai 2 + 


a 2 2 ) 




a?/ai 


(2) 






+ 2 0102 V2 




o 2 /( V2 ai + 02) 


(3) 






+ axd^Vs 




(2oi + 02V3)/o 2 


(4) 






+ ai02 




026/(01 + «2C) 


(5) 






* 




02/01 


(6) 






— 2 ai02& 




a 2 c/(— Oi + 026) 


(7) 






+ 2 oiO«6 




(01 + o 2 6)/o2C 


(8) 






+ 2 01026 




(01 + 02b) /(— a 2 b) 


(9) 
(10) 


i 




+ 2 01O26 




(Oi + 02C)/(— O26) 

Oi/o 2 


(11) 


4 




+ 2 0102 sin (5? 


-so 


(O16 + 02c)/(oiCi — a 2 6 2 ) 



16 ANSWERS TO PROBLEMS 

(12) (a x 2 + a 2 2 ) + 2 a x a 2 cos (5 2 — Si) (— a x c — a 2 c)/(a\b + 026) 

(13) " -2aia 2 6 (a^ + a 2 b)/(- a x b + a 2 c) 

(14) " +2 aia 2 cos 2 8 c(ai -f « 2 )/b(ai — a 2 ) 

(15) " +2 aia 2 cos 2 5 c(ai + a 2 ) /6(a 2 — a 2 ) 

Pages 385-386 

1. (a) 2. V/P ; (6) 2. y/ 2q2 +^ 2 ; (C ) 2. \7^P ; 



sin a 
27" 



— • 4. 7024 ft. 5. p , _ p . 



(d) 2, \/ a -f 

* g 

7. 2tt l/^L • - , or, 2tt ^y - • 8. 12.5 sec. per hr. 10. 4000 miles. 

12. (a) (b) (c) (d) (e) (/) 

P. 2^/ZJI, 2,1/^-, 2xl/*> 2x\/^, *rl/*\7l, ^V/-~V7^- 
V V3 g y V3g * Y £ MV5 MV3 

n * l « JL 4/2 4/2 

2V3' 2V3' 2' V2' a V5 a ' V3' 

Page 403 

1. 0.013P. 2. 4*\/^ 4 - 2,tV/^P- 5. 2r\/|A^?. 
v ▼ 15 gr * 5 



LIBRARY OF CONGRESS 




